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Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = {x = Rn | Añ = 0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁v₁ + ··· + tk√k, where t1,..., tk are free parameters, then the vectors 1,..., Uk form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then {ã € Rª | Añ = 0} = {ã € R″ | A'ñ = 0}. Deduce that if B is the row-reduced echelon form of A, then {ã € R” | Añ = 0} = {ã € R″ | Bì = 0}. (b) Looking at the row-reduced echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x, is a leading variable, then we have n xi = Σ Gijli 1 j=i+1 xj free for some constants Cij E R. Let i₁ < ... < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system BÃ = Ō are given by x1 Ti1-1 Ti1 Ti+1 1 = = t1 ++tk Tik-1 Tik Tik+1 In Uk √1 Prove that the vectors 1,..., Vk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix 1 2 3 4 A=0 0 1 2 00 2 4

Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = {x = Rn | Añ = 0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁v₁ + ··· + tk√k, where t1,..., tk are free parameters, then the vectors 1,..., Uk form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then {ã € Rª | Añ = 0} = {ã € R″ | A'ñ = 0}. Deduce that if B is the row-reduced echelon form of A, then {ã € R” | Añ = 0} = {ã € R″ | Bì = 0}. (b) Looking at the row-reduced echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x, is a leading variable, then we have n xi = Σ Gijli 1 j=i+1 xj free for some constants Cij E R. Let i₁ < ... < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system BÃ = Ō are given by x1 Ti1-1 Ti1 Ti+1 1 = = t1 ++tk Tik-1 Tik Tik+1 In Uk √1 Prove that the vectors 1,..., Vk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix 1 2 3 4 A=0 0 1 2 00 2 4

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = {x = Rn | Añ = 0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁v₁ + ··· + tk√k, where t1,..., tk are free parameters, then the vectors 1,..., Uk form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then {ã € Rª | Añ = 0} = {ã € R″ | A'ñ = 0}. Deduce that if B is the row-reduced echelon form of A, then {ã € R” | Añ = 0} = {ã € R″ | Bì = 0}. (b) Looking at the row-reduced echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x, is a leading variable, then we have n xi = Σ Gijli 1 j=i+1 xj free for some constants Cij E R. Let i₁ < ... < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system BÃ = Ō are given by x1 Ti1-1 Ti1 Ti+1 1 = = t1 ++tk Tik-1 Tik Tik+1 In Uk √1 Prove that the vectors 1,..., Vk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix 1 2 3 4 A=0 0 1 2 00 2 4
Transcribed Image Text:Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = {x = Rn | Añ = 0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁v₁ + ··· + tk√k, where t1,..., tk are free parameters, then the vectors 1,..., Uk form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then {ã € Rª | Añ = 0} = {ã € R″ | A'ñ = 0}. Deduce that if B is the row-reduced echelon form of A, then {ã € R” | Añ = 0} = {ã € R″ | Bì = 0}. (b) Looking at the row-reduced echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x, is a leading variable, then we have n xi = Σ Gijli 1 j=i+1 xj free for some constants Cij E R. Let i₁ < ... < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system BÃ = Ō are given by x1 Ti1-1 Ti1 Ti+1 1 = = t1 ++tk Tik-1 Tik Tik+1 In Uk √1 Prove that the vectors 1,..., Vk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix 1 2 3 4 A=0 0 1 2 00 2 4
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